4190. Prime Palindromes 一亿以内的质数回文数

Description

The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 100,000,000); both a and b are considered to be within the range .

Input

There are multiple test cases.

Each case contains two integers, a and b.

a=b=0 indicates the end of input.

Output

For each test case, output the list of palindromic primes in numerical order, one per line.

 

Sample Input

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5 500
0 0

Sample Output

5
7
11
101
131
151
181
191
313
353
373
383



题目链接：4190. Prime Palindromes
题意分析：求给定区间内的质数回文数
题目分析：
1.多组测试数据，所以先打表。
2.先求质数再判断回文，效率低下；所以先构造回文数，再判断质数。
3.偶数位的回文数都能被11整除，自己证明去。所以，偶数位的回文数除了11都是合数。
4.一个k位数，可以构造出一个奇数位的回文数。比如13，可以构造131；189可以构造18981.所以100000000内的只要从1构造到9999即可。
5.若范围为1000000000，那么明显超出int范围，要用long long。当然此题无此陷阱。

//http://soj.me/show_problem.php?pid=4190
#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

vector<int> v;

bool prime(int x)
{
    for (int i = 2; i < sqrt(x+0.5); i++)
        if (x%i == 0)
            return 0;
        
    return 1;
}

void get()
{
    // 11是唯一的偶数位的质数回文数。
    v.push_back(11);
    
    //构造奇数位的回文数
    for (int i = 2; i <= 10000; ++i)
    {
        int tmp = i/10, sum;
        
        for (sum = i; tmp != 0; tmp /= 10)
        {
            sum = sum*10 + tmp%10;
        }
        
        if (prime(sum))
            v.push_back(sum);
    }
}
int main()
{
    get();
    sort(v.begin(), v.end());    //因为是不按顺序push，所以要sort
    int a, b;
    while(cin >> a >> b, a||b)
    {
        for (int i = 0; i < v.size(); ++i)
        {
            if (v[i] >= a)
            {
                if (v[i] > b)
                    break;
                
                printf("%d\n", v[i]);
            }
        }
    }
}

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